2019

MPROVEMENT

Mechanical Engineering

Theory of Machines
Answer Key of Objective & Conventional Questions
 1 Mechanisms and Machines

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1. (c) 14. (6)

2. (c) 15. (1)

3. (a) 16. (c)

4. (a) 17. (1)

5. (c) 19. (c)

6. (a) 20. (c)

7. (d) 21. (b)

8. (d) 22. (a)

9. (d) 23. (b)

10. (a) 24. (b)

11. (c) 25. (1)

12. (d) 26. (a)

13. (d) 27. (180)

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 Rank Improvement Workbook 3

Solution : 28
In this mechanism

95.7°
50 70
b c

B D
a 20 A
70 d
(a)

μ = 95.7°
C

°
77.6
b c

A D
a d

B (c)
μ = 45.6°

60°
b c

B
a

A d D
(d)

μ = 60°
And the input angle, θ = 60°

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 Mechanical Engineering •
4 Theory of Machine

Solution : 29
The mechanism has three sub-chains:
(i) ABC, a slider-crank chain
(ii) ABDE, a four-bar chain
(iii) AEFG, a four-bar chain
(DEF is a locked chain as it has only three links.)
• As the length BC is more than the length AB plus the offset of 2 units, AB acts as a crank and can
revolve about A.
• In the chain ABDE,
Length of the longest link = 8, Length of the shortest link = 4, Length of the other links = 8 and 6
Since 8 + 4 < 8 + 6, it belongs to the class-I mechanism. In this case as the shortest link is fixed, it is a
double-crank mechanism and thus EF and AG can revolve fully.
• In the chain AEFG,
Length of the longest link = 8, Length of the shortest link = 4, Length of the other links = 6 and 6
Since 8 + 4 = 6 + 6, it belongs to the class-I mechanism. In this case as the shortest link is fixed, it is a
double-crank mechanism and thus EF and AG can revolve fully.
AS DEF is a locked chain with three links, the link EF revolves with the revolving of ED. With the revolving
of ED, AG also revolves.
Solution : 30
(a) The mechanism has a sliding pair. Therefore, its degree of freedom must be found from Gruebler’s
criterion. Total number of links = 8
7
4
5 1
3 6

2
8
1
1
(At the slider, one sliding pair and two turning pairs)
F = 3(N – 1) – 2P1 – P2
= 3(8 – 1) – 2 × 10 – 0 = 1
Thus, it is a mechanism with a single degree of freedom.
(b) The system has a redundant degree of freedom as the rod of the mechanism can slide without
causing any movement in the rest of the mechanism.
∴ Eeffective degree of freedom = 3(N – 1) – 2P1 – P2 – Fr
= 3 (4 – 1) – 2 × 4 – 0 – 1 = 0
As the effective degree of freedom is zero, it is a locked system.

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 2 Velocity & Acceleration
Analysis

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1. (b) 12. (67) (66.5 to 67.5)

2. (0.33) 13. (a)

3. (d) 14. (301.59)

4. (c) 15. (2.74)

5. (b) 16. (a)

6. (d) 17. (3)

7. (0) 18. (1000)

8. (b) 19. (45)

9. (1) 20. (2.69)

10. (b) 21. (8.386)

11. (c) 22. (a)

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 Mechanical Engineering •
6 Theory of Machine

Solution : 23
vd = gd = 0.56 m/s
ωbq = 5.63 rad/s counter-clockwise
ωba = 6.3 rad/s counter-clockwise
Velocity of rubbing at the crank pin
B = 0.0268 m/s
Solution : 24
v s = 0.276 m/s
vpq = 0.177 m/s
ωrs = 0.279 rad/s clockwise

Solution : 25
vd = 2.28 m/s

Solution : 26
αcd = 33.25 rad/s2
Solution : 27
aA/0 = 3.507 × 103 mm/s2
VB = 2.3 × 4 = 9.2 m/s
aB = 2.6 × 105 m/s2
α= 6666.67 rad/s2
Acceleration of mid point aG = 2800 m/s2
Solution : 28
By measurement VR = vector O1r = 1.61 m/s
ωDO2 = 1.112 rad/s anticlockwise about O2
Solution : 29
Length of crank = 25 cm
V = 4.1887 m/s

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 3 Cams

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1. (c) 13. (2.37)

2. (c) 14. (720)

3. (c) 15. (0.628)

4. (a) 16. (1.95)

5. (a) 17. (d)

6. (c) 18. (b)

7. (c) 19. (d)

8. (c) 20. (d)

9. (c) 21. (c)

11. (c) 22. (a)

12. (b) 23. (c)

24. (b)

25. (a)

26. (b)

27. (b)

28. (a)

29. (d)

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 Mechanical Engineering •
8 Theory of Machine

Solution : 30
vmax = 226.3 mm/s
fmax = 3.413 m/s2
During descent vmax = 216 mm/s
fmax = f = 0
Solution : 31
During descent vmax = 824.7 mm/s
vmax = 549.8 mm/s
fmax = 38862 mm/s2 = 38.882 m/s2
fmax = 17272 mm/s2 or 17.272 m/s2
Solution : 32

30 mm
0
φa δ1 φd
0′ 1′ 2′ (a)
2.52
1 2 3′
Velocity

3
4 4′
120° (m/s)
5
5′ (b) 2.64
30
90°

6 464.6
°

12′ 12
Acceleration

11
11′ 6 211.9
10 9 8 7 6′
10′
9′
8′ 6′ 211.9
7′
2
(m/s )
464.6
Solution : 33

1′ O′
2′
1
3′ 2
3 14 14′
4′ δ2 13 13′
4 φa

5 12
12′
δ1 φd
5′ 6 11
11′
6 10
9
7 8
6′ 10′

9′
6′
7′ 8′

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 Rank Improvement Workbook 9

vmax = 360 m/s

fmax = 4320 mm/s2 or 4.32 m/s2

Solution : 34
x = 40 ω2 cos θ
N = 609.9 rpm

Solution : 35
The displacement diagram for the given flat reciprocating follower movement will be as:
6 6′ 6′
5
5′ 7′
4 8′
4′

3
3′ 9′
2′ 10′
2
1′ 11′
12′
1
1 2 3 4 5 6 6 7 8 9 10 11 12
Dwell Dwell
Outstroke Return stroke
(φa) (δ1) (δd) (δ2)

Given: φa = 120°, h(lift) = 20 mm, δ1 = 30°, φd = 120°, δ2 = 90°

Motion is SHM both during outward and inward stroke, minimum radius of cam (rc) = 25 mm.
Construction:
25
1. First draw the displacement diagram now construct the cam
a
profile as follows b
1
2. Draw a circle with radius (rc = 25 mm) c 2
3. Take angles (φa, δ1, φd and δ2) in the counter clockwise 3

90
0°

°
d

12
direction if the cam rotation is assumed clockwise 4

4. Divide φa and φd into same number of parts as in the 5

displacement diagram. (Example take 6 equal parts) e
30
6 0° 12 l
12 °
5. Draw radial lines (0-1, 0-2, 0-3,, etc. ....) 11
f 7
6. On the radial lines produced, take distances equal to the 8 10 k
9
lift of the follower beyond the circumference of the circle
with radius rc, i.e., 1 - 1′, 2 - 2′, 3 - 3′, etc. g
j
7. Draw the follower in all the positions by drawing i
perpendiculars to the radial lines at 1′, 2′, 3′, etc. In all the h

positions, the axis of the follower passes through centre O

8. Draw a curve tangential to the flat faces of the follower representing the cam profile.

Solution : 36
Radial component of cam force is given by;
Fr = 61 N
Torque = 0.651 N.m
Solution : 37
In this motion:
(V0)max = 1.2 m/s
a = 72 m/s2

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 4 Gear and Gear Train

20. (a)

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21. (a)
1. (c)
22. (c)
2. (7.48)
23. (0.86)
3. (91)
24. (10.29)
4. (360)
25. (9.04)
5. (a)
26. (c)
6. (c)
27. (b)
7. (b)
28. (a)
8. (d)
29. (c)
9. (d)
30. (b)
10. (d)
31. (c)
11. (c)
32. (b)
12. (39207.076)
33. (b)

34. (a)

35. (d)
13. (c)
36. (c)
14. (c)
37. (d)
15. (b)
38. (70)
16. (a)
39. (d)
17. (b)
40. (18)
18. (b)
41. (c)
19. (b) „„„„

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 Rank Improvement Workbook 11

Solution : 42
Velocity of sliding = 57049 mm/min = 950.8 mm/s
Maximum velocity of sliding = 1017.1 mm/s
Solution : 43
Addendum of the wheel = 8.3 mm

ra2 = 30271
Addendum of the pinion = 14 mm
Arc of contact = 58.2 mm
Solution : 44
Length of arc of contact = 30.788 mm
n = 1.6334
Angle of action by the pinion, θp = 0.54 radian
θp = 30.95°
(a) = 0.388
(b) = 0.348
(c) = 0
Solution : 45
T = 49.44
n = 1.78
Solution : 46
Addendum, a = 0.8010 m
Stubbing required = 19.9% or 20%
Solution : 47
TF = 72
TS = 18
Speed of P, NP = – 166.67 rpm
Therefore, speed of planet Gear P is 166.67 rpm in opposite direction to S and A.

Solution : 48
Speed of output shaft = – 50 rpm (clockwise)
Speed of output shaft = 39.5 rpm (clockwise)

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 5 Flywheel and Governors

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1. (c) 18. (d)

2. (d) 19. (323.6)

3. (d) 20. (991.735)

4. (d) 21. (183.33)

5. (243.17) 22. (22360)

6. (0.04) 23. (0.38)

7. (0.38) 24. (d)

8. (d) 25. (b)

9. (a) 26. (0.51)

10. (d) 27. (31.98)

11 (206.04) 28. (0.0444)

12. (b) 29. (51.3)

13. (a) 30. (31.4)(30 to 33)

14. (a) 31. (b)

15. (c) 32. (c)

16. (d) 33. (c)

17. (a) 34. (c)

35. (a)

36. (d)

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 Rank Improvement Workbook 13

Solution : 37
m = 1217.4 kg
Solution : 38
I = 293.3 kg.m2

Solution : 39
Power of motor = 4276.5 Watt = 4.28 kW
Solution : 40
Motor power = 0.3 kW
M = 988.68 kg
Solution : 41
N = 167 rpm
Range of speed = 4.163 rpm
Solution : 42
m = 5.2 kg
s = 32.72 N/mm
compression of the spring = 33.2 mm
Solution : 43
I = 1394.58 kg-m2
Solution : 44
K = 0.678%
Solution : 45
N = 430.43 rpm
Solution : 46
Power of engine = 261.8 kW
I = 785.166 kg-m2
α = 2.547 rad/sec2
Solution : 47
Initial compression,S1 = 10.03 cm
x1 = 11.65 cm

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Balancing and Gyroscope

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„„„„
(d)

(a)

(a)

(a)

(a)

(a)
(c)
10.

11.

12.

13.

14.

15.

16.
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6
(d)

(b)

(d)

(b)
(a)

(a)

(a)
(c)

(c)
1.

2.

3.

4.

5.

6.

7.

8.

9.
 Rank Improvement Workbook 15

Solution : 17
v = 151 km/h

Solution : 18
mc1 = 3.13 kg at 253°
mc1 = 3.14 kg
Solution : 19
ma = 17.37 kg
θa = 294.6° or 294°36′
ld = –309 mm
lb = – 376 mm
lc = –126 mm

Solution : 20
m4 = 178.7 kg
θ4 = 248.2°
m1 = 178.7 kg = m4
θ1 = 201.8°
Swaying couple = 3030.3 N.m
Variation in tractive force = 10100 N
Balance mass for reciprocating parts only = 74.46 kg
Maximum pressure on rails = 45326 N
Minimum pressure on rails = 23344 N
Velocity of wheels = 88.36 km/h
Solution : 21
m3 = 448 kg
m2 = 438 kg
Solution : 22
R1 = 4431.8 N
R2 = 8223.8 N
R3 = 2567.2 N
R4 = 6359.2 N
R1 = 8158.2 N
R2 = 4366.2 N
R3 = 6426.8 N
R4 = 2632.8 N

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 Mechanical Engineering •
16 Theory of Machine

Solution : 23
Reaction at bearing, B = 98.6 N
Reaction at bearing, A = 59.4 N
Solution : 24
mA = 9.67 kg, mD = 7.89 kg
Angular position of the mass at D = 252.7° (w.r.t. B)

Solution : 25
m = 92.8 kg
θ = 201.48°
Solution : 26
Resultant = 8224.6 N
M = 40 kg

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 7 Vibrations

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1. (d) 13. (b)

2. (d) 14. (a)

3. (d) 15. (1.11)

4. (a) 16. (c)

5. (a) 17. (b)

6. (3) 18. (2.28)

7. (b) 19. (0.56)

8. (d) 20. (26.74)

9. (d) 21. (1.9052)

10. (d) 22. (a)

11. (b) 23. (0.667)

12.. (a) 24. (1)

25. (8)

26. (2)

27. (0.05)

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 Mechanical Engineering •
18 Theory of Machine

Solution : 28
1
 + aρg × 2 xx = 0
alρ× 2 xx
2
2g
x + x =0
l
Solution : 29
A = 0.00298 m = 2.98 mm
Solution : 30
fn = 2.85 Hz
Solution : 31
δ= 0.693
C= 45.809 Nm/rad
periodic time of oscillation = 1.503 × 10–3 sec
fn = 669.2 Hz
Solution : 32
2
k1l12 + mgl ⎛ Cl2 ⎞
ωd = −⎜
ml 2 ⎝ 2ml 2 ⎟⎠
Solution : 33
Nc = 2598 r.p.m
Solution : 34
C = 400.824 N/m/s
fd
= 0.99
fn
Td = 0.32 sec
Solution : 35
δ = 0.405
Damping coefficient, C = 32.745 N.m.s/rad
ωd = 4233.72 rad/s
Td = 1.484 × 10–3 s
Solution : 36
FT = 38.6 N
FT = 367 N
Amplitude of the forced vibration of the machine at resonance = 8.7 mm
Solution : 37
C = 2970 N-m/rad
Solution : 38
A = 130 mm
φ = 42.4897°
Solution : 39
k = 1.607 N/mm

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